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505-555 (Easy)|   Geometry|      
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Bunuel
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A cylindrical pitcher with a radius 5 is filled with beer to a height 06 Sep 2018, 00:57
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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84% (01:15) correct 16% (01:59) wrong based on 44 sessions

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A cylindrical pitcher with a radius 5 is filled with beer to a height of 9 inches. If the beer is poured from the first pitcher to a second pitcher with no spillage and fills the second pitcher to a height of 4 inches, what is the radius of the second pitcher?


A. 15 inches
B. 12.5 inches
C. 9 inches
D. 8 inches
E. 7.5 inches
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E
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A cylindrical pitcher with a radius 5 is filled with beer to a height 06 Sep 2018, 03:15
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Bunuel
A cylindrical pitcher with a radius 5 is filled with beer to a height of 9 inches. If the beer is poured from the first pitcher to a second pitcher with no spillage and fills the second pitcher to a height of 4 inches, what is the radius of the second pitcher?


A. 15 inches
B. 12.5 inches
C. 9 inches
D. 8 inches
E. 7.5 inches
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Volume of beer in first pitcher = Volume of Beer in second pitcher

Volume of a cylinder = \(πr^2h\)

i.e. \(π*5^2*9 = π*r^2*4\)
i.e. \(r = 7.5\)

Answer: option E
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A cylindrical pitcher with a radius 5 is filled with beer to a height 06 Sep 2018, 08:06
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Bunuel
A cylindrical pitcher with a radius 5 is filled with beer to a height of 9 inches. If the beer is poured from the first pitcher to a second pitcher with no spillage and fills the second pitcher to a height of 4 inches, what is the radius of the second pitcher?


A. 15 inches
B. 12.5 inches
C. 9 inches
D. 8 inches
E. 7.5 inches
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\(πR^2H = πr^2h\)

Or, \(π*5^2*9 = π*r^2*4\)

Or, \(5^2*3^2 = r^2*2^2\)

Or, \(5*3 = 2r\)

So, \(r = 7.5\), ANswer must be (E)


PS : IMHO, the highlighted part must be with a radius 5 inches Bunuel
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A cylindrical pitcher with a radius 5 is filled with beer to a height 06 Sep 2018, 11:34
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Bunuel
A cylindrical pitcher with a radius 5 is filled with beer to a height of 9 inches. If the beer is poured from the first pitcher to a second pitcher with no spillage and fills the second pitcher to a height of 4 inches, what is the radius of the second pitcher?


A. 15 inches
B. 12.5 inches
C. 9 inches
D. 8 inches
E. 7.5 inches
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The volume of beer is equal in both pitchers.

Volume of [the liquid in] a cylinder, and thus in the first pitcher:
\(V =\pi r^2 h\)
\(V=\pi* 5^2*9=225\pi\)

That volume of beer is in the second pitcher. Solve for radius. Second pitcher:
\(225\pi=\pi r^2h=\pi* r^2* 4\)
\(\frac{225}{4}=r^2\)
\(56.25=r^2\)
\(\sqrt{56.25}=r\)

Between which two perfect squares does \(56.25\) lie?

\(7^2=49 < 56.25 < 64=8^2\)

The radius is between \(7\) and \(8\).

Answer E

Alternatively, check the answer choices.
(C) \(9^2=81\) -> too great. We need 56.25
(D) \(8^2=64\) -> too great.

By POE, (E) is the answer.
And: (E) 7.5 is between 7 and 8
\(7^2 = 49\)
\(8^2=64\)
\(49<56.25<64\) - That works.

Answer E
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